HW8Soln

# HW8Soln - Solutions to HW8(due 4/2 4/3 Page 274 7 dy dx =...

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Unformatted text preview: Solutions to HW8 (due 4/2, 4/3) Page 274 7: dy dx = dy/dt dx/dt = e x + y y- 1 = e x e y y- 1 . This is separable, leading to Z ( y- 1) e- y dy = Z e x dx + C, or- ye- y = e x + C. or e x + ye- y = C The mixture of y and e- y means we cannot solve explicitly for y in terms of x , but, we can solve for x in terms of y , x = ln (- ye- y- C ) 8: dy dx = dy/dt dx/dt = 3 x 2- 2 xy x 2- 2 y- 3 This is not separable or linear, so we check to see if it is exact. First rewrite in the form (3 x 2- 2 xy ) dx + (- x 2 + 2 y- 3 ) dy = 0 If this is Mdx + Ndy , then ∂M/∂y =- 2 x and ∂N/∂x =- 2 x , so the equation is exact. To find F ( x, y ) such that ∂F/∂x = M , we integrate with respect to x (holding y constant). We get that F ( x, y ) = x 3- x 2 y + f ( y ) for some function f ( y ) depending only on y . Taking the partial with respect to y and setting it equal to N we get- x 2 + f ( y ) =- x 2 + 2 y- 3 so that f ( y ) = 2 y- 3 and f ( y ) =- y- 2 + C . Therefore we can take F ( x, y ) = x 3- x 2 y- y- 2 so that the integral curves are of the form x 3- x 2 y- y- 2 = C There is no nice explicit form for this. 9: dy dx = dy/dt dx/dt = e x + y 2 y- x This is not separable or linear, so we check to see if it is exact. First rewrite in the form ( e x + y ) dx + ( x- 2 y ) dy = 0 . If this is Mdx + Ndy , then ∂M/∂y = 1 and ∂N/∂x = 1, so the equation is exact. To find F ( x, y ) such that ∂F/∂x = M , we integrate with respect to x (holding y constant). We get that F ( x, y ) = e x + xy + f ( y ) for some function f ( y ) depending only on y . Taking the partial with respect to y and setting it equal to N we get x + f ( y ) = x- 2 y so that f ( y ) =- 2 y and f ( y ) =- y 2 + C . Therefore we can take F ( x, y ) = e x + xy- y 2 so that the integral curves are of the form e x + xy- y 2 = C . This can be solved for y using the quadratic formula, y = x ± p x 2 + 4( C- e x ) 2 , 1 but in most applications, that form is not more desirable than the implicit form. 10: To find critical points, solve x 2- 1 = 0 , xy = 0. The first equation implies x = ± 1. Combining that with the second equation we get that y = 0. Thus there are two critical points, (1 , 0), and (- 1 , 0). The phase plane equation is dy dx = dy/dt dx/dt = xy x 2- 1 which is separable leading to Z dy y = Z x dx x 2- 1 + C....
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HW8Soln - Solutions to HW8(due 4/2 4/3 Page 274 7 dy dx =...

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